Question

13.23, решите срочно​

Answer 1

Ответ:

$x^4\, y''+x^3\, y'=4\ \ \Big|:x^4\ne 0\ ,\ \ \ \ \ \ \ \ y(1)=1\ ,\ y'(1)=-2\\\\y''+\dfrac{1}{x}\cdot y'=4\ \ ,\ \ \ \ y'=p(x)\ \ ,\ \ y''=p'(x)\ \ \Rightarrow \\\\\\p'+\dfrac{1}{x}\cdot p=4\ \ \ (linejnoe)\\\\p=uv\ ,\ p'=u'v =uv'\\\\u'v+uv'+\dfrac{1}{x}\cdot uv=4\ \ ,\ \ \ u'v+u\cdot (v'+\dfrac{1}{x}\cdot v)=4\\\\a)\ \ v'+\dfrac{1}{x}\cdot v=0\ \ ,\ \ \dfrac{dv}{dx}=-\dfrac{v}{x}\ \ ,\ \ \int \dfrac{dv}{v}=-\int \dfrac{dx}{x}\ \ ,\\\\ln|v|=-ln|x|\ \ ,\ \ \ v=x^{-1}\ \ ,\ \ \ v=\dfrac{1}{x}$

$b)\ \ u'v=4\ \ ,\ \ \dfrac{du}{dx}\cdot \dfrac{1}{x}=4\ \ ,\ \ \ \int du=4\int x\cdot dx\ \ ,\ \ \ u=4\cdot \dfrac{x^2}{2}+C_1\ \ ,\\\\u=2x^2+C_1\\\\c)\ \ p(x)=uv=\dfrac{1}{x}\cdot (2x^2+C_1)\ \ \Rightarrow \ \ \ y'=\dfrac{1}{x}\cdot (2x^2+C_1)\ \ ,\ \ \ \dfrac{dy}{dx}=\dfrac{2x^2+C_1}{x}\ ,\\\\\int dy=\int (2x+\dfrac{C_1}{x}\ )\, dx\ \ ,\ \ \ y=2\cdot \dfrac{x^2}{2}+C_1\cdot ln|x|+C_2\\\\\\\boxed{\ y_{obshee}=x^2+C_1\, ln|x|+C_2\ }$

$d)\ \ y(1)=1\ \ \to \ \ \ 1=1^2+C_1\cdot \underbrace{ln1}_{0}+C_2\ \ ,\ \ 1+C_2=1\ \ ,\ \ C_2=0\\\\y'(1)=-2\ \ ,\ \ y'(x)=\dfrac{1}{x}\cdot (2x^2+C_1)\ \ \to \ \ \ -2=1\cdot (2+C_1)\ \ ,\ \ C_1=-4\\\\\\\boxed{\ y_{chastnoe}=x^2-4\cdot ln|\, x\, |\ }$